2016.3.1 Question 1

Notice that

1. Let $x+a=\sqrt{b-a^2}\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u$. When $x\to -\infty$, $u\to -\frac{\pi }{2}$, and when $x\to +\infty$, $u\to \frac{\pi }{2}$. We have also

   $$\begin{array}{llll}\hfill dx=d(x+a)& =d\sqrt{b-a^2}\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u& \hfill & \\ \hfill & =\sqrt{b-a^2}d\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u& \hfill & \\ \hfill & =\sqrt{b-a^2}{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u<!--\; FUNCTION\; APPLICATION\; -->du.& \hfill & \end{array}$$

   Therefore, we have

   $$\begin{array}{llll}\hfill I_1& ={\int \; <!--\; nolimits\; -->}_{-\infty }^{+\infty }\frac{dx}{{(x+a)}^2+(b-a^2)}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\sqrt{b-a^2}{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u<!--\; FUNCTION\; APPLICATION\; -->du}{{\left(\sqrt{b-a^2}\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u\right)}^2+(b-a^2)}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\sqrt{b-a^2}{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u<!--\; FUNCTION\; APPLICATION\; -->du}{(b-a^2)({\text{tan}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u+1)}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u<!--\; FUNCTION\; APPLICATION\; -->du}{\sqrt{b-a^2}{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{\sqrt{b-a^2}}& \hfill & \\ \hfill & =\frac{\pi }{\sqrt{b-a^2}},& \hfill & \end{array}$$

   as desired.

2. Using the same substitution, we have

   $$\begin{array}{llll}\hfill I_n& ={\int \; <!--\; nolimits\; -->}_{-\infty }^{+\infty }\frac{dx}{{[{(x+a)}^2+(b-a^2)]}^n}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\sqrt{b-a^2}{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u<!--\; FUNCTION\; APPLICATION\; -->du}{{\left[(b-a^2){\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u\right]}^n}& \hfill & \\ \hfill & =\frac{1}{\sqrt{b-a^2}}{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\left[(b-a^2){\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u\right]}^{n-1}}.& \hfill & \end{array}$$

   Therefore,

   $$2n(b-a^2)I_{n+1}=(2n-1)I_n,$$

   is equivalent to

   $$2n\sqrt{b-a^2}{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\left[(b-a^2){\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u\right]}^n}=(2n-1)\frac{1}{\sqrt{b-a^2}}{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\left[(b-a^2){\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u\right]}^{n-1}}$$

   is equivalent to

   $$2n(b-a^2){\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\left[(b-a^2){\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u\right]}^n}=(2n-1){\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\left[(b-a^2){\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u\right]}^{n-1}}$$

   is equivalent to

   $$2n{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u}=(2n-1){\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n-2}u}.$$

   Notice that

   $$\begin{array}{llll}\hfill {\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n-2}u}& ={\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u<!--\; FUNCTION\; APPLICATION\; -->du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{d\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u}& \hfill & \\ \hfill & =\underset{\begin{array}{c}a\to \frac{\pi }{2}\\ b\to -\frac{\pi }{2}\end{array}}{\text{lim}<!--\; FUNCTION\; APPLICATION\; -->}{\left[\frac{\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u}\right]}_b^a-{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u<!--\; FUNCTION\; APPLICATION\; -->d{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{-2n}u& \hfill & \\ \hfill & =\underset{\begin{array}{c}a\to \frac{\pi }{2}\\ b\to -\frac{\pi }{2}\end{array}}{\text{lim}<!--\; FUNCTION\; APPLICATION\; -->}{\left[\text{sin}<!--\; FUNCTION\; APPLICATION\; -->u{\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^{2n-1}u\right]}_b^a-{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-2n\text{sec}<!--\; FUNCTION\; APPLICATION\; -->u\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{-2n-1}u\text{tan}<!--\; FUNCTION\; APPLICATION\; -->u<!--\; FUNCTION\; APPLICATION\; -->du& \hfill & \\ \hfill & =2n{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\text{tan}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u<!--\; FUNCTION\; APPLICATION\; -->du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u}& \hfill & \\ \hfill & =2n{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{({\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}u-1)<!--\; FUNCTION\; APPLICATION\; -->du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u}& \hfill & \\ \hfill & =2n{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n-2}u}-2n{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u}.& \hfill & \\ \hfill & & \hfill \end{array}$$

   This means

   $$(2n-1){\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n-2}u}=2n{\int \; <!--\; nolimits\; -->}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{du}{{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2n}u},$$

   which is exactly what was desired.

3. Proof by induction:

   • Base Case. When $n=1$,

     $$\text{LHS}=I_1=\frac{\pi }{\sqrt{b-a^2}},$$

     $$\text{RHS}=\frac{\pi }{2^{2\cdot 1-2}{(b-a^2)}^{1-\frac{1}{2}}}\left(\genfrac{}{}{0.0pt}{}{2\cdot 1-2}{1-1}\right)=\frac{\pi }{\sqrt{b-a^2}}\left(\genfrac{}{}{0.0pt}{}{0}{0}\right)=\frac{\pi }{\sqrt{b-a^2}}.$$

   • Induction Hypothesis. Assume for some $n=k\in \mathbb{N}$, we have

     $$I_n=\frac{\pi }{2^{2n-2}{(b-a^2)}^{n-\frac{1}{2}}}\left(\genfrac{}{}{0.0pt}{}{2n-2}{n-1}\right).$$

   • Induction Step. When $n=k+1$,

     $$\begin{array}{llll}\hfill I_n& =I_{k+1}& \hfill & \\ \hfill & =\frac{2k+1}{2(k+1)(b-a^2)}{I}_k& \hfill & \\ \hfill & =\frac{2k+1}{2(k+1)(b-a^2)}\cdot \frac{\pi }{2^{2k-2}{(b-a^2)}^{k-\frac{1}{2}}}\left(\genfrac{}{}{0.0pt}{}{2k-2}{k-1}\right)& \hfill & \\ \hfill & =\frac{\pi }{2^{2k}{(b-a^2)}^{k+\frac{1}{2}}}\frac{(2k-2)!}{(k-1)!(k-1)!}\frac{(2k+1)(2k+2)}{{(k+1)}^2}& \hfill & \\ \hfill & =\frac{\pi }{2^{2k}{(b-a^2)}^{k+\frac{1}{2}}}\frac{2k!}{k!k!}& \hfill & \\ \hfill & =\frac{\pi }{2^{2k}{(b-a^2)}^{k+\frac{1}{2}}}\left(\genfrac{}{}{0.0pt}{}{2k}{k}\right)& \hfill & \\ \hfill & =\frac{\pi }{2^{2n-2}{(b-a^2)}^{n-\frac{1}{2}}}\left(\genfrac{}{}{0.0pt}{}{2n-2}{n-1}\right).& \hfill & \end{array}$$

   Therefore, by the principle of mathematical induction, for $n\in \mathbb{N}$,

   $$I_n=\frac{\pi }{2^{2n-2}{(b-a^2)}^{n-\frac{1}{2}}}\left(\genfrac{}{}{0.0pt}{}{2n-2}{n-1}\right),$$

   as desired.